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Sujay
You are a farmer, and you are trying to make the most efficient use of your resources. You have invested enough capital for a barn house which is 40 feet by 70 feet (its base) and 100 feet of fencing for your animals. You decide to have an outdoor lot for your animals. Your goal is to have the lot with the most area with the least amount of fencing (optimize). The lot must be rectangular, and one side of the rectangular may be a wall from the barn. A picture will be posted below.
The area of your pen is the length times the width. Since you are going to have three sides of the pen enclosed by your fencing (the last side will be from the barn), we have a second specification that must be met, the length plus twice the width is 100. Rearrange the second equation to solve for a single variable (either l or w) and substitute into the first equation. You will get a quadratic function; find the maximum value, and you have the width (or maybe length) that corresponds to the maximum area. One more substitution gives the dimensions of the pen..
\[A=lw\]\[100=l+2w \implies l=100-2w \implies\]\[A=lw=A(w)=(100-2w)w=-2w^2+100\]To find the maximum (or minimum in other cases) of a quadratic function f(x)=ax^2+bx+c, we take the value of the function at x=-b/(2a). In this case, that gives us\[w=\frac{-100}{2(-2)}=25 \implies l=50\]by substitution. So our enclosure is 25' by 50', against the side of the barn that is 70' long.
Wow thanks, I totally forgot we don't need calculus for this problem.
Can anyone do it using optimization?
The optimization problem is the same, except when you get A(w), you take the derivative and set it to zero. Since the second derivative is negative at that value of w, the function has a maximum at that point.\[A(w)=−2w^2+100w \implies A'(w)=-4w+100=0 \implies w=25\]\[A"(w)=-4<0 \implies maximum~at~critical~value\]